I was reading Gilberg Trudinger chapter 9 Page 221.
Contact sets: If u is arbitrary continuous function on $\Omega$ we define the upper contact set of u denoted by $\Gamma^+$ or $\Gamma_u^+$ to be the subsets of $\Omega$ where graph of $u$ lies below a support of hyperplane in $\mathbb R^{n+1}$ i.e. \begin{equation}\label{eq:95} \Gamma^+=\{y\in \Omega:u(x)\le u(y)+p(x-y)\forall x\in \Omega, p=p(y)\in \mathbb R^n\}. \end{equation}
Clearly , if u is concave function iff $\Gamma^+=\Omega$. When $u\in C^1(\Omega)$ we must have $p=Du(y)$ in above equation as any support hyperplane must be tangent hyperplane to graph of u. Furthermore , when $u\in C^2(\Omega)$, the Hessian matrix $D^2u$ is non positive on $\Gamma^+$. In general set $\Gamma^+$ is relatively closed to $\Omega$.
Normal mapping : For arbitrary function $u\in C^0(\Omega)$ we define the normal mapping ,$\chi (y)=\chi_u(y)$ of a point $y\in \Omega$ to be set of slopes of support hyperplanes at y lying above the graph i.e \begin{equation}\label{eq:96} \chi(y)=\{p\in \mathbb R^n: u(x)\le u(y)+p(x-y)\qquad \forall x\in \Omega\}. \end{equation}
Clearly $\chi(y)\neq \varphi$ iff $y\in \Gamma^+$. Further more when $u\in C^1(\Omega)$, then $\chi(y)=Du(y)$ on $\Gamma^+$.
Now till now I think I understand . Now Author given example of normal mapping of non differentiable functions.
Let $\Omega=B=B_R(z)$ and $u$ is function whose graph is cone with base $\Omega$ and vertex $(z,a)$ for some positive $a\in \mathbb R$.
$$ u(x)=a\left(1-\frac{|x-z|}{R}\right). $$ Then we have $$ \chi_u(y)=\left\{\begin{matrix} \frac{-a(y-z)}{R|y-z|} & y\ne z \\ B_{a/R}(0) &y=z \end{matrix}\right. $$
I think as at $y=z$ function is nondifferetiable . So $p$ can be any direction . So collectively $\chi(z)=B_{a/R}(0)$. But I do not understand why particular radius choose?
Also I do not understand normal mapping at $y\ne z$. I thought it would vector in direction along tangent of cone.
Any Help will be greatly appreciated.